证明: $$lim_{n o+infty}cos^{n}left(frac{1}{x} ight)dx=0$$
证明：作变量替换 $u=frac{1}{x}$, 则有
egin{align*}
int_{1}^{infty}left|frac{cos^{n}u}{u^{2}}
ight| du&=int_{1}^{frac{pi}{2}}left|frac{cos^{n}u}{u^{2}}
ight| du+sum_{k=0}^{infty}int_{left(k+frac{1}{2}
ight)pi}^{left(k+frac{3}{2}
ight)pi}left|frac{cos^{n}u}{u^{2}}
ight| du\
&leq cos^{n} 1 int_{1}^{frac{pi}{2}}frac{1}{u^{2}}du+frac{4}{pi^{2}}sum_{k=0}^{infty}frac{1}{(2k+1)^{2}}int_{0}^{pi}|cos^{n}u|du\
&=cos^{n} 1 int_{1}^{frac{pi}{2}}frac{1}{u^{2}}du+frac{1}{2}int_{0}^{pi}|cos^{n}u|du
end{align*}
令$n o infty$,知 $cos^{n} 1 o 0, int_{0}^{pi}|cos^{n}u|du o 0$. 这是由于
$$ int_{0}^{pi}|cos^{n}u|du=2int_{0}^{frac{pi}{2}}cos ^{n}u du=2 I_{n}$$
由于单调有界收敛原理知$n o infty$极限必然存在. 所以考虑偶序列递推公式和 Wallis 公式
$$I_{2n}=frac{(2n-1)!!}{(2n)!!}frac{pi}{2}simsqrt{frac{pi}{2}}sqrt{frac{1}{2n}},\, n o infty$$